From the Fourier Theorem applied to $h$ you know $c_n$ in terms of $h$. Now all you're doing is rewriting that so that you learn what the $c_n$ are in terms of $g$

The way you did it obscures things slighty. Instead of doing the change of variables and then substituting in $g$ for something, you could have swapped $h$ for $g$ at the beginning. Then the change of variables is obvious so that you get $g(\nu)$ and not $g$ with some more complicated argument

You don't have to apply the same change of variables in both places. It isn't logically required that you do this rewriting.

@KevinDriscoll Ok, we're rewriting $c_n$ in terms of $g$. I understand we did that. But what confuses me is that, to achieve that, we have to change the interval of integration. So, in my head, a voice is saying that we need to change back to the original variable of the integral, i.e. $t$.

Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of real numbers with the following property. There exists a real number $0 < r < 1$ and an integer $N_0$ such that $\forall n \geq N_0$, $|a_{n}-a_{n-1}| \leq r|a_{n-1}-a_{n-2}|$. Prove that $(a_n)$ convergs.

Now, since we're working in $\mathbb{R}$, it suffices to show that this sequence is Cauchy. Beyond that, I've tried the triangle inequality, and that hasn't helped - so I'm kinda stuck.

@KevinDriscoll Why don't we have to change the current variable of the integral (i.e. $v$) back to $t$ (which means that we would also change the interval of integration, which means that $c_n$ wouldn't anymore look like $d_n$)?

@nbro I think you're getting confused because in the problem statement they wrote $c_n$ as an integral over $t$. And then they wrote $d_n$ also as an integral over $t$. These $t$ are not the same.

$t$ is just a dummy variable. It parameterizes the integral over which the function is periodic. We could have called it $Trump$ and done the $c_n$ over $t$ from $-\pi, \pi$ and then done the $d_n$ interval over $Trump$ where $Trump \in [0, N]$

I was not realizing that changing the variables is only a way to eventually more conveniently solve, in this case, an integral over the same interval of the original function, even though the interval of the integral after the changes of variables has changed.

@nbro I can understand the confusion, mathematically speaking. In the abstract, it isn't super clear. But if we think about the mathematics as being a representation of some feature of our universe, then it becomes more clear that what we're talking about can be thought of as different ways of describing the same thing

@nbro It still doesn't look like so, because, from what I know, once we calculated an integral with the new variable and we obtained a result, we need to re-express that result in terms of the original variable...

So we can re-express $u$ in terms of $x$ and then plug the approriate $x$ values into the anti-derivative in terms of $x$. Or we can figure out what $u$ values then $x$-endpoints correspond to. And then we can plug those $u$ values directly into the anti-derivative in terms of $u$. Same thing

Side comment: A lot of people seem not to be taught (especially in high school?) to rewrite the new integral with new limits of integration, but do the substitution, do the integral, and then rewrite everything in terms of the original variables and then evaluate. Ugh.

@KevinDriscoll So, in my particular case, you're saying that the final integral we obtain can then be re-evaluated using the original variable $t$ (and with the corresponding interval) or evaluated using $v$, correct?

But I took calc 1 at a community college that was relatively rigorous about actually teach math and science for engineers because they send lots of people to engineering schools after 2 years of community college

@TedShifrin THIS ANNOYS ME OT NO END! Students come to me and they're like "How do I show this limit?" And it takes me freakin 10 minutes to figure out how to do it using their tools because I would do it in 30 seconds with Taylor series

Anyway, thanks a lot guys for all your valuable and precious help! I've not taken a serious nap since the last WW. I probably could have solved this alone (by searching around and revising a few concepts) if I wasn't so tired.........

@nbro I invite you to do a simple example, like the square wave periodic on $[0,N]$ and teh corresponding re-scaled and re-centered version of the square wave periodic on $[-\pi, \pi]$ to see that in fact the sequence of constants that you get for one are exactly the same as the sequence of constants for the other. The thing that changes is which harmonic (AKA which exponential) they go with. But the sequence of constants is identical.

@brot Yes, my mistake. I assumed the sequence was monotone because of the $m<n$ bit. What you can also say is. Suppose $a_n$ is bounded by $M$. Consider $[-M,M]\in\mathbb{R}$. Note that the number of elements of $\{a_n\}$ is limited if we are to preserve the condition of "spacing" between any two elements of the sequence.

Question: If a person is providing an answer that is assuming something which is not necessarily being assumed in the initial question, like "whole numbers are being used," for example, is that a potential issue?

> What do the Real Numbers Really Look Like? What does the real line look like if you look at it really close up? We think of it as a continuum of points extending indefinitely in two directions, but what if you could look at it under a microscope and were able to turn up the magnification higher and higher. What would you begin to see? You might be disappointed since you will never get to a stage where you would see, “one rational number, three irrational numbers, one rational number, ... “ The real numbers are self-similar, they “look alike” no matter what the scale. So how do we “visuali…

Well... until you decided to use an uncountable magnification, then suddenly everything turned to discrete points (there are infinitesimal gaps between any two reals in the hyperreals)

If countable infinity is the notion of the point you should get to if you go arbitrarily close to it in steps, then uncountable infinities are the notions you should get to if the aforementioned approaching becomes a one step process

That's one way to visualise $\omega_1$ despite its inherent lack of describable structure:

> You zoomed pass points so fast that any points that take countably many steps to reach takes only one step, and you use this newfound speed to try to get as close as you can to the next possible limit point

So, any uncomputable functions with growth rate comparable to $\omega_1$ should exceed any countable uncomputable and computable function right at the first step

(To be proved...)

and that's the hallmark of the notion of uncountability: It is something so big that no countable process can reach it

I mean, let $g,h$ be uncomputable functions where $g$ is comparable to the fast growing hierarchy $f_{\omega_1}$ while $h$ is comparable to the fast growing hierarchy $f_{\alpha}$ where $\alpha < \omega_1$. Then I suspect $g(1) > h(1)$ now matter what $h$ is

Do we expect the fast growing hierarchy to be able to go beyond $\omega_1$ because of issue of $\omega_1$ is it lacks a fundamental sequence, so there will be trouble in expanding $f_{\omega_1}$?

Because from what I learnt, fast growing hierarchy $f_{\alpha} > f_{\beta}$ if $\alpha > \beta$

Since $\omega_1 > \alpha$ where $\alpha$ is countable, I should expect the fast growing hierarchy should follow similarly, but I might be wrong because it might be possible it breaks down after $\omega_1^{CK}$

(To others who may care) Anyway, the above is a minor digress as currently I am reading about the nested interval theorem to remind myself on how to prove the compactness of closed intervals in the reals

Hmm... so the outline of the proof combining these sources is as follows:

1. The reals are linearly ordered. Pick a closed interval $[a,b]$. 2. Generate a sequence of intervals by picking some $c \in [a,b]$ and $d \in [a,b]$ such that $c \leq d$

3. All bounded sequences will converge to its infimum or supremum. Therefore these are contained in any interval constructed

4. In the case of creating nested intervals by halving, the infimum of the sequence with 1st term a and supremum of the sequence with 1st term b coincides, to some $y$. Since [y,y] is be definition a singleton, we have the required result

Bolzano-Weierstrass Theorem

1. Pick a [a,b]

2. Generate nested intervals by halving a and b

3. By nested interval, the arbitrary intersection exists and is some number y

4. Let U be a neighbourhood of y defined by $(y-r,s-y)$. Pick one of the closed intervals I form the nested intervals such that $I\subset U$. Regardless of $U$ there are uncountably many points in these intervals, hence y is an accumulation point. Since [a,b] is arbitrary, it followed every real number is an accumulation point

Proving [a,b] is compact

... and I don't want to use contradiction...

1. Let $C$ be a cover of $[a,b]$

2. Any such $C$ must obey the property that $\{a,b\} \subset C \cap [a,b]$

3. Since the arbitrary union of any open set is open, it follows there exists no sequence of unions of open intervals with infimum a or supremum b that can contain a or b.

4. Now consider some open interval $(c,d)$ such that $c < a, d > b$. $(c,d)$ can be covered by some infinite cover $C$.

5. By Bolzano-Weierstrass Theorem and that arbitrary union of open intervals are open, all points on the real line are accumulation points, therefore $C$ cannot consists of a subsequence of open intervals with some supremum or infimum $x \in [a,b]$ that includes $x$. (In more intuitive terms, the open cover cannot contains an infinite "tail" converging to some point x in [a,b] without leaving x uncovered)

6. Therefore there are no countable sequence of open intervals within $[a,b]$, and thus any infinite sequence in the infinite cover $C$ must be located outside $[a,b]$

7. Therefore the union of open intervals within $[a,b]$ is finite. Meanwhile for the region $(c,e)$ and $(f,d)$ where $e < a,b > f$, the infinite bounded sequence of intervals can form a union to give finite open intervals as a sub cover. for said region.

Let $A$ be a subset of $R$ which consist of $0$ and the numbers $\frac{1}{n}$, for $n=1,2,3,\dots$. I want to prove that $K$ is compact directly from the definition of compact.
So, given any open cover of $A$, I should be able to find a finite subcover. Proving a set is compact is much difficult...

Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of real numbers with the following property. There exists a real number $0 < r < 1$ and an integer $N_0$ such that $\forall n \geq N_0$, $|a_{n}-a_{n-1}| \leq r|a_{n-1}-a_{n-2}|$. Prove that $(a_n)$ convergs.

Strange vectors and matrices inspired from the notion of datasets and data structures:

Let some dataset be the space $D=\Bbb{R}^3 \otimes \Bbb{R}^2 \otimes \Bbb{R}^2$. For example, the 3D vector space is a vector field of wind velocity, and the 2D vector space are precipitation with time and temperature with time

We can consider some vector $v \in D$ the represent a datapoint, given by the 3 tuple $(a,b,c)$ where $a \in \Bbb{R}^3, b,c \in \Bbb{R}^2$

Now, suppose we are interested in the wind velocity in the x direction, and the average temperature, precipitation between Monday and Thursday. This becomes an operator $\phi : D \to \Bbb{R}^3 \otimes \Bbb{R} \otimes \Bbb{R}$

Therefore $\phi (v)$ is an element wise inner product within each vector space, given by:

More generally, we can think of linear operators where one of the dimensions are irregular

thus we can have something like a stack of matrices in the space $\bigotimes_{i \in I}\Bbb{R}^i$ where $I=\{1,2\times 2,3 \times 3,4 \times 4,3 \times 3,2 \times 2,1\}$

We can also consider the continuum case of a function of operators given by the space $\prod_{i \in J}\phi_i$ where $J=[a,b]$

In such scenario, the input are functions and the output are also functions and each $\phi_i$ can have a different domain and range

For example, let $f$ be a real function. Then $\phi (f)$ is also a function. Thus $\phi$ acts as a in general, nonlinear and not necessary continuous map on the space of all functions

Also unlike usual nonlinear functionals, $\phi$ can have a very complicated shaped domain due to the domains and ranges of all $\phi_i$ are not necessary the same

As my real and functional analysis get better, I will try to investigate $\phi$ in more details. But for now, I am still trying to figure out how to prove the Lebesgue outer measure of a closed interval is indeed closed

I have the Bolzano-Weierstrass Theorem for real closed intervals proved, and I am trying to prove the Lebesgue measure of closed intervals without having the expression b-a appearing anywhere except the conclusion of the proof

I also have the compactness of [a,b] proved.

But using the definition of Lebesgue outer measure, I am not sure how I can get the notion on how the measure of an open interval is derived, and that is needed to do the proof of the closed interval case

It might seemed like a stupid question or that I overlooked something, but I am trying to justify why the lebesgue outer measure of open intervals (a,b) is b-a

Well, anyway, my goal is to solve the equation numerically using GMRES. I've done some numerical experiments that suggest it works and I know there are some papers on this. I just realized I've been referring to this equation as fredholm for the past couple weeks and that's not quite right

1. There is a maximum and a minimum 2. We start at the minimum and the maximum can be reached I.e. the maximum is a successor of some number 3. We also require every number is a successor of some number, except the minimum 4. There are only successors and the minimum, and between any two successor there are no numbers

5. Some extra condition to rule out something like:

0,1,2,3,4,...,5,6,7,8,9

Meanwhile I am still have trouble figuring out $\lambda^*(a,b)$

How to prove the Lebesgue outer measure of an open interval (a,b) is b-a?

actually, there's one thing that seems basic but I am not terribly sure. Since measure theory aim to generalise the notion of length and size, is the notion of length of an interval defined before we have a measure?

(question left for later) whether for a generic measure $\mu$ there's always some set choose measure is defined instead of calculated and we use these as building blocks to compute the measure of other sets in the $\sigma$-algebra.

Thanks, I think I should have all the ingredient I need to prove $\lambda^*[a,b]=b-a$ now. Having done the proofs on the compactness of closed intervals in reals, the nested interval theorem and the Bolzano-Weierstrass Theorem for closed intervals, I should have everything to do the above proof using these and topological arguments. Will write out the proof afte…

Proof: $\{a\} = [a,a]$ is compact since it is a closed interval in the reals. Any open cover of $\{a\}$ will contain the point $a$. For every open cover $C$, there's always an open interval $(b,c)$ where $b<a<c$ that is contained in $C$. The infimum of $(b,c)$ tends to zero as $b,c$ get arbitirarily close to $a$ thus the sequence of nested open intervals converges to $a$. Therefore the natural measure of $\{a\}$ is $\lim_{c\to a}c - \lim_{b\to a} b = a-a=0$. Thus any singleton has zero Lebesgue measure

Therefore the Lebesgue outer measure of $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ follows from subadditivity of the measure. Hence $\lambda^*[a,b] \leq 0+b-a+0=b-a$ and equality is ensured because the union is disjoint

Now for rationals: $\Bbb{Q} \cap [a,b]$ :

Proof: The rationals are countable. Let $f$ be an enumeration of rationals given by the sequence under the Cantor pairing function $\{\frac{a+b}{2},\frac{a+b}{2}+1,\frac{a+b}{2}-1,\frac{a+b}{2}+\frac{1}{2}, \frac{a+b}{2} -\frac{1}{2},\frac{a+b}{2}+2,\frac{a+b}{2}-2,...\}$

if $\frac{a+b}{2}$ is rational, this sequence give all the rationals in $[a,b]$. Thus we can then take the union of each rational here and then use countable subadditivity to establish $\lambda^*(\Bbb{Q}\cap [a,b])=\sum_{i=1}^{\aleph_0}0 = 0$

if $\frac{a+b}{2}$ is irrational, then (thinking...)

If $\frac{a+b}{2}$ is irrational, pick any rational $p > \frac{a+b}{2}$ and then use $f$ to generate a sequence. Once again, all rationals in $[a,b]$ will be enumerated and thus the proof proceed analogously

that is, I want to think of the computation entirely in terms of covers

I guess the first step is to work out whether $\Bbb{I}\cap [a,b]$ is compact, or even countably compact

A cover $C$ can be countable by having countably many open intervals converging to some rational. Since there are countably many rationals, there are at least countably many such sequences of open intervals that converges to them (1 or 2 more if the endpoints a or b are rationals). Hence such $C$ form a countable cover